This could have been written clearer. I think the culprit is the section:
The problem was first identified over a century ago. At the time, mathematicians knew that “the real numbers are bigger than the natural numbers, but not how much bigger. Is it the next biggest size, or is there a size in between?” said Maryanthe Malliaris of the University of Chicago, co-author of the new work along with Saharon Shelah of the Hebrew University of Jerusalem and Rutgers University.
In their new work, Malliaris and Shelah resolve a related 70-year-old question about whether one infinity (call it p) is smaller than another infinity (call it t). They proved the two are in fact equal, much to the surprise of mathematicians.
If read quickly, this suggests that $\mathfrak{p}$ and $\mathfrak{t}$ refer to the cardinality of the set of reals and the set of naturals, respectively. This is not the case, though.
So what sort of thing are $\mathfrak{p}$ and $\mathfrak{t}$, then?
$\mathfrak{p}$ and $\mathfrak{t}$ are what are known as cardinal characteristics of the continuum (CCCs) - cardinals which are (i) known to be uncountable, and (ii) measure how big a set of reals has to be to have some "universality" property.
For example, one simple CCC is the dominating number, $\mathfrak{d}$: this is the smallest cardinality of a set $F$ of functions $\mathbb{N}\rightarrow\mathbb{N}$ such that for each $g:\mathbb{N}\rightarrow\mathbb{N}$ there is some $f\in F$ such that $f(n)>g(n)$ for all but finitely many $n$ (we say $f$ dominates $g$). Clearly $\mathfrak{d}$ is at most continuum (since that's how many functions $\mathbb{N}\rightarrow\mathbb{N}$ there are in the first place), and it's also uncountable: if $f_i:\mathbb{N}\rightarrow\mathbb{N}$ for $i\in\mathbb{N}$, the function $$h(i)=\sum_{j\le i}f_j(i)=f_1(i)+f_2(i)+...+f_i(i)$$ is not dominated by any of the $f_i$s.
Another simple CCC is the bounding number, $\mathfrak{b}$. This is "dual" to $\mathfrak{d}$ (in a sense that can be made precise): $\mathfrak{b}$ is the smallest size of any family $G$ of functions $\mathbb{N}\rightarrow\mathbb{N}$ such that no single $f$ dominates all functions in $G$. Again, $\mathfrak{b}$ is clearly at most continuum, and is uncountable since any countably many functions can be dominated by a single function (think about the construction of the $h$ above).
Now cardinal arithmetic is notoriously badly behaved - even basic facts about it tend to be undecidable in ZFC. For example, ZFC doesn't even prove that $\kappa<\lambda \implies 2^\kappa<2^\lambda$. So it's really exciting to see ZFC-provable facts about infinite cardinalities; conversely, it's important to understand when certain questions can't be resolved in ZFC alone. In this context, what we care about is comparing CCCs. We can think about it this way: the two trivial CCCs are $\omega_1$ ("the smallest size of an uncountable set of reals") and $2^{\aleph_0}$ ("the smallest size of a set containing all the reals"); and in between we have the interesting CCCs. Of course, if $\omega_1=2^{\aleph_0}$ then the whole picture collapses; this is the continuum hypothesis, and it's consistent with ZFC. At the far other end, it's known that we can separate certain CCCs - e.g. that it is consistent with ZFC that $\mathfrak{b}<\mathfrak{d}$. (An interesting topic is separating multiple CCCs simultaneously - see this MO question.)
This leaves open:
What equalities between CCCs can we prove in ZFC? What inequalities can we disprove?
As an example of the latter, ZFC proves that $\mathfrak{b}\le\mathfrak{d}$ - we can't have $\mathfrak{b}>\mathfrak{d}$ (this is a good exercise). More broadly, the collection of disprovable inequalities between many CCCs (not including $\mathfrak{p}$ and $\mathfrak{t}$, though) is summed up in Cichon's diagram. Malliaris and Shelah proved a result of the former kind - showing that two CCCs were in fact equal. My understanding is that this type of result is much, much rarer even in general, and of course in this particular case it was extremely surprising (see Shelah's quote in the linked article).
Of course, I haven't tried to define $\mathfrak{p}$ and $\mathfrak{t}$; the definitions are there, but they're a bit technical, and more to the point it's hard to see why someone would care. A good analysis of them is not something I can fit into an MSE answer; but hopefully what I've written explains a bit about where this sort of thing can come from!
(Let $\mathbb I$ be the notation for the irrationals).
This is actually a map from $f:\mathbb R \to \mathbb I$ and not the other way, so the fact that your don't ever get $2$ (or any other rational number) is a feature, not a bug.
Of course this means the there must be an inverse from $f^{-1}:\mathbb I \to \mathbb R$. This wold be. If $x = q + \sqrt 2*n; n\ge 1; n\in \mathbb Z$ then map $x \to q + (n-1)\sqrt 2$; otherwise map all things to themselves. Thus we can get $2$ via $2 + \sqrt 2 \to 2$.
And this seems weird because .... what does the $\sqrt 2$ have to do with anything. And as you probably guessed, and irrational $\omega$ will do as well.
So to demonstrate:
1) $f: \mathbb R \to \mathbb I$.
If $q \in \mathbb Q$ then $q = q + 0*\omega$ so $f(q) = q + \omega \in \mathbb I$.
If $x = q + k\omega; q\in \mathbb Q; k \in \mathbb Z; k \ge 1$ then $f(x) = q + (k+1) \omega\in \mathbb I$.
And if neither of those cases $f(x) = x \in \mathbb I$.
So $f: \mathbb R \to \mathbb I$.
2) $f$ is unto.
If $x \in \mathbb I$ then either $x = q + k\omega$ where $q $ is rational and $k$ is a positive number or .... $x$ doesn't.
If $x$ does then $f(x - \omega) = f(q + (k-1)\omega) = q + k\omega = x$.
And if $x$ doesn't then $f(x) = x$.
So either way there exist a real number that maps to $x$.
So $f$ is onto.
3) $f$ is one-to- one.
Let $f(w) = x \in \mathbb I$. If $x \ne q + k*\omega$ for any rational $q$ and positive integer $z$ then if $w = q + n*\omega$ for some rational $q$ and natural $n$ then $f(w) = q + (n+1)\omega \ne x$ so that's impossible. So $f(w) = w = x$. That's the only one possible value that maps to $x$.
If $x = q + k \omega$ for some rational $q$ and positive integer $k$ and $w$ does not equal any $r + n*\omega$ for any rational $r$ not any rational $n$ then $f(w) = w \ne q+k\omega = x$ so that's impossible. So $w = r + n*\omega$. So $f(w) = r + (n+1)\omega = q + k\omega$.
So $r-q = (k- n-1)\omega$. But if $k-n-1\ne 0$ then the RHS is irrational and that is impossible. So $k - n - 1 = 0$ and $n = k-1$ and $r = q$ and $w = q + (k-1)\omega$ and that is the only one possible value that maps to $x$.
So $f$ is one-to-one.
... and so $f$ is a bijection between $\mathbb R\to \mathbb I$.
====
Okay, this feels ... wierd. But why?
What's really going on is
$\mathbb I = \mathbb R \setminus \mathbb Q$ and $|\mathbb R| >|\mathbb Q|=$ countable. When we want to make a bijection from a set to itself minus a countable subset of itself the "trick" is to leave the vast majority of the subset alone but to shove the "chain" of countable members to a countable number of exceptions that just follow an infinite but countable chain out of the way. So we "shove" all rational $q$ to $q + \omega$ That obvioulsy works. But were do we shove all the $q + \omega$ to? Well, to $q + 2\omega$ of course! But where do we shove those to? Well, we just shove all the $q + k\omega$ to $q + (k+1)\omega$.
Hopefully seen like that (albeit informal and handwavy) it is clearer why id does (and must) work.
Best Answer
If on the right you have $0.01010101 ...$ and so on ad infinitum, on the left you have an infinite number $10101010....$ which is not on your list because an element of the natural numbers is itself not an "infinite number".
So all you can create on the RHS are finitely terminating rationals which is a countable set.